Lemma 37.29.2. Let $f : X \to Y$, $s : Y \to X$ be as in Situation 37.29.1. If $g : Y' \to Y$ is any morphism, consider the base change diagram
so that we obtain $(X')^0 \subset X'$. Then $(X')^0 = (g')^{-1}(X^0)$.
Proof. Let $y' \in Y'$ with image $y \in Y$. We may think of $X^0_ y$ as a closed subscheme of $X_ y$, see for example Morphisms, Definition 29.26.3. As $s(y) \in X^0_ y$ we conclude from Varieties, Lemma 33.7.14 that $X_ y^0$ is a geometrically connected scheme over $\kappa (y)$. Hence $X_ y^0 \times _ y y' \to X'_{y'}$ is a connected closed subscheme which contains $s'(y')$. Thus $X_ y^0 \times _ y y' \subset (X'_{y'})^0$. The other inclusion $X_ y^0 \times _ y y' \supset (X'_{y'})^0$ is clear as the image of $(X'_{y'})^0$ in $X_ y$ is a connected subset of $X_ y$ which contains $s(y)$. $\square$
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)